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\title{Chapter 16: Direct Images}
\author{SCC ET AL}
%\institute[XX大学]{XX大学\quad 数学与统计学院\quad 数学与应用数学专业}
%\date{2025年6月}

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% 封面页
\begin{frame}
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% 目录页
\begin{frame}{Contents}
  \tableofcontents
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% Section 0
%\section{INTRO.}
\begin{frame}{intro. }
    
In the previous chapter we saw that starting with an $A_m$-module and a polynomial map $F: R^n \to K^m$ we can construct an $A_n$-module, its inverse image. 

We shall now study a similar construction, which uses $F$ to associate an $A_m$-module to an $A_n$-module, its direct image. 

Curiously, the direct image is easier to define for right modules; and it is with them that we start. 

Throughout this chapter, the conventions of 14.1.2 remain in force.

\end{frame}

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% Section 1
\section{Right Modules}
\begin{frame}[allowframebreaks]{A. }

Let us briefly recall the definition of the inverse image. 

Let $F: X \to Y$ be a polynomial map. 

Let $M$ be a left $A_m$-module. 

The {\color{red}inverse image} of $M$ under $F$ is $F{\,}^* M = K[X] \otimes_{K[Y]} M$. 

This is a $K[X]$-module. 

It becomes an $A_n$-module with the $\partial_{x_i}$ acting according to the formula

\begin{equation}
\partial_{x_i} (h \otimes u) = \frac{\partial h}{\partial x_i} \otimes u + \sum_{j=1}^{m} h \frac{\partial F_j}{\partial x_i} \otimes \partial_{y_j} u.
\end{equation}

Let us rewrite this definition in a slightly different way. 

Since $A_m \otimes_{A_m} M \cong M$, we have that
\begin{equation}
F{\,}^* M \cong K[X] \otimes_{K[Y]} A_m \otimes_{A_m} M.
\end{equation}

Writing $D_{X \to Y}$ for $F{\,}^* A_m = K[X] \otimes_{K[Y]} A_m$, one has that
\begin{equation}
F{\,}^* M = D_{X \to Y} \otimes_{A_m} M.
\end{equation}

Note that $D_{X \to Y}$ is a left $A_n$-module and a right $A_m$-module. 

One easily checks that these two structures are compatible and that it is in fact an $A_n$-$A_m$-bimodule.

{\color{blue}The bimodule $D_{X \to Y}$ is the key to the direct image construction. }

Let $N$ be a right $A_n$-module. 

The tensor product
\begin{equation}
F_* N = N \otimes_{A_n} D_{X \to Y}
\end{equation}
is a right $A_m$-module, which is called the {\color{red}direct image} of $N$ under the polynomial map $F$.

Consider the projection $\pi: X \times Y \to Y$ defined by $\pi(X, Y) = Y$. 

From Ch.14, §3, it follows that
\begin{equation}
\pi^* A_m \cong K[X] \widehat{\otimes} A_m.
\end{equation}

As a left $A_n$-module, $K[X]$ is isomorphic to $A_n / \sum_{1}^{n} A_n \partial_{x_i}$. 

Hence, by Corollary 13.2.3,
\begin{equation}
\pi^* A_m \cong A_{m+n} / \sum_{1}^{n} A_{m+n} \partial_{x_i}
\end{equation}
is an isomorphism of $A_{m+n}$-$A_m$-bimodules. 

Now if $N$ is a right $A_{m+n}$-module, then by Corollary 12.4.7,
\begin{equation}
\pi_* N \cong N / \sum_{1}^{n} N \partial_{x_i}
\end{equation}
as right $A_m$-modules.

\end{frame}

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% Section 2
\section{Transposition}
\begin{frame}[allowframebreaks]{B. }

We will now see how one can turn right modules into left modules, and vice versa. 

That will allow us to do direct images for left modules. 

We present the construction for general algebras.

Let $R$ be a $K$-algebra. 

A {\color{red}transposition} of $R$ is an isomorphism $\tau: R \to R$ of the underlying $K$-vector space that satisfies
\begin{enumerate}
    \item $\tau(ab) = \tau(b)\tau(a)$,
    \item $\tau^2 = \tau$.
\end{enumerate}

Examples of transpositions abound. 

The most familiar occurs in the matrix ring over a field. 

An example in the Weyl algebra is provided by the map $\tau: A_n \to A_n$ defined by
\begin{equation}
\tau(h \partial^\alpha) = (-1)^{|\alpha|} \partial^\alpha h,
\end{equation}
where $h \in K[x_1, \ldots, x_n]$ and $\alpha \in \mathbb{N}^n$. 

We will refer to it as the {\color{red}standard transposition} of $A_n$. 

If $d \in A_n$, we also use the notation $d^\tau$ for $\tau(d)$.

Let us go back to the general construction. 

Let $R$ be a $K$-algebra and $\tau$ a transposition of $R$. 

If $N$ is a right $R$-module then we define a left $R$-module $N^t$ as follows. 

As an abelian group, $N^t = N$. 

If $a \in R$ and $u \in N^t$ then the left action of $a$ on $u$ is defined by $a \diamond u = u\tau(a)$. 

The left action $\diamond$ is called the {\color{red}transposed action}. 

Instead of checking in detail that $N^t$ is a left $R$-module, let us prove just one typical property, namely
\begin{equation}
(ab) \diamond u = a \diamond (b \diamond u)
\end{equation}
where $a, b \in R$ and $u \in N$. 

The left hand side is, by definition, $u\tau(ab)$. 

Since $\tau$ is a transposition, it equals $u\tau(b)\tau(a)$. 

But this is $a \diamond (b \diamond u)$.

The same construction can be used to turn left modules into right modules. 

Indeed, let $M$ be a left $R$-module. 

The right $R$-module $M{\,}^t$ equals $M$ as an abelian group.

The right action of $a \in R$ on $u \in M$ is defined by $u \diamond a = \tau(a)u$. 

Note that since $\tau^2 = \tau$, the two constructions above are inverse to each other. 

If $N$ is a right $R$-module, then $(N^t)^t = N$; and the same holds for left modules.

{\color{blue}It is very easy to transpose the action when the module is cyclic. }

Let $R$ be a $K$-algebra with a transposition $\tau$. 

If $\mathcal{J}$ is a left ideal of $R$, define $\mathcal{J}^t = \{\tau(a): a \in \mathcal{J}\}$. 

This is a right ideal of $R$.


\textbf{Proposition.}

Let $R$ be a $K$-algebra with a transposition $\tau$ and $\mathcal{J}$ a left ideal of $R$. 

The right $R$-module $(R/\mathcal{J})^t$ is isomorphic to $R/\mathcal{J}^t$.

\textbf{Proof:} If $a \in R$, denote by $\overline{a}$ its image in $(R/\mathcal{J})^t$. 

Consider the map
\begin{equation}
\phi : R \to (R/\mathcal{J})^t
\end{equation}
defined by $\phi(a) = \overline{\tau(a)}$. 

It is clearly a homomorphism of additive groups. 

If $a, b \in R$, then
\begin{equation}
\phi(ab) = \overline{\tau(ab)},
\end{equation}
which equals $\overline{\tau(b)\tau(a)} = \phi(a) \diamond b$. 

Thus $\phi$ is a homomorphism of right $R$-modules. 

One can easily check that it is surjective; let us calculate its kernel. 

Suppose that $a \in R$ satisfies $\phi(a) = 0$. 

Then $\tau(a) \in \mathcal{J}$ or, equivalently, $a \in \mathcal{J}^t$. 

Hence $\ker \phi \subseteq \mathcal{J}^t$. 

The other inclusion is similarly proved, and we conclude that $\ker \phi = \mathcal{J}^t$. 

By the first homomorphism theorem, we have that $R/\mathcal{J}^t \cong (R/\mathcal{J})^t$.

Let us illustrate the proposition with a concrete example of a module over the Weyl algebra. 

Let $\mathcal{J} = A_n x_n$, and let $\tau$ be the standard transposition of $A_n$. 

Since $\tau(x_n) = x_n$, it follows that $\mathcal{J}^t = x_n A_n$. 

Thus
\begin{equation}
(A_n / \mathcal{J})^t \cong A_n / x_n A_n.
\end{equation}

This example will surface again later on.

All this can be extended to bimodules. 

Let $R_1$, $R_2$ be $K$-algebras with transpositions $\tau_1$, $\tau_2$, respectively. 

Let $M$ be an $R_1$-$R_2$-bimodule. 

The {\color{red}transposed bimodule} $M{\,}^t$ is obtained by applying the above constructions to both the left and the right actions on $M$. 

Thus $M{\,}^t$ and $M$ have the same underlying abelian groups and, if $a_1 \in R_1$, $a_2 \in R_2$ and $u \in M$, then
\begin{equation}
a_2 \diamond u \diamond a_1 = (a_1)^{\tau_1} u (a_2)^{\tau_2}.
\end{equation}

These actions are compatible with each other, since the original actions were so. 

Thus $M{\,}^t$ is an $R_2$-$R_1$-bimodule.


The transposition of bimodules over the Weyl algebra has its own peculiarities. 

Suppose that $m \leq n$ and that $\tau_m$ and $\tau_n$ are the standard transpositions of $A_m$ and $A_n$, respectively. 

Then $A_m \subseteq A_n$ and $\tau_n$ restricted to $A_m$ equals $\tau_m$. 

This considerably simplifies the calculations. 

For instance, if $J$ is a left ideal of $A_n$ which is a right $A_m$-module, then it follows from these considerations and Proposition 2.1 that $(A_n / J)^t$ is isomorphic to $A_n / J^t$ as an $A_m$-$A_n$-bimodule. 

The left ideal $J = A_n x_n$ is an example of this situation. 

This also means that we may denote the standard transposition simply by $\tau$, irrespective of the index of the Weyl algebra in question.

For future reference, we must consider the behaviour of the transposition under tensor product.

\textbf{Lemma.} 

Let $R_1$, $R_2$ and $R_3$ be $K$-algebras with transposition. 

Suppose that $M_1$ is an $R_1$-$R_2$-bimodule and $M_2$ an $R_2$-$R_3$-bimodule. Then
\begin{equation}
(M_1 \otimes_{R_2} M_2)^t \cong M_2{\,}^t \otimes_{R_2} M_1{\,}^t
\end{equation}
as $R_1$-$R_3$-bimodules.


\textbf{Proof:} Consider the map
\begin{equation}
\phi : M_2{\,}^t \times M_1{\,}^t \to (M_1 \otimes_{R_2} M_2)^t,
\end{equation}
defined by $\phi(u_2, u_1) = u_1 \otimes u_2$, where $u_i \in M_i$. 

We show that $\phi$ is balanced, and leave the proof that it is bilinear to the reader. 

If $a \in R_2$ and $\tau_2$ is the transposition of $R_2$, then
\begin{equation}
\phi(u_2 \diamond a, u_1) = \phi(\tau(a) u_2, u_1).
\end{equation}

By the definition of $\phi$,
\begin{equation}
\phi(u_2 \diamond a, u_1) = u_1 \otimes \tau(a) u_2.
\end{equation}

Now, using that the tensor product $M_1 \otimes_{R_2} M_2$ is balanced, we get that
\begin{equation}
\phi(u_2 \diamond a, u_1) = u_1 \tau(a) \otimes u_2.
\end{equation}

Hence,
\begin{equation}
\phi(u_2 \diamond a, u_1) = \phi(u_2, u_1 \tau(a)) = \phi(u_2, a \diamond u_1).
\end{equation}

Thus $\phi$ induces a homomorphism of $R_1$-$R_3$-bimodules,
\begin{equation}
\overline{\phi} : (M_1 \otimes_{R_2} M_2)^t \to M_2{\,}^t \otimes_{R_2} M_1{\,}^t.
\end{equation}

It is clear from the definition of $\overline{\phi}$ that it is bijective, and the proof is complete.


\end{frame}

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% Section 3
\section{Left Modules}
\begin{frame}[allowframebreaks]{C. }


Using the results of §2, we will do direct images for left modules. 

Let $F: X \to Y$ be a polynomial map. 

Consider the $A_n$-$A_m$-bimodule $D_{X \to Y}$. 

{\color{blue}
Using the standard transposition for $A_m$ and $A_n$ defined in the previous section, put
\begin{equation}
D_{Y \leftarrow X} = \left(D_{X \to Y}\right)^t.
\end{equation}

This is an $A_m$-$A_n$-bimodule.
}

Let $M$ be a left $A_n$-module. 

The {\color{red}direct image} of $M$ by $F$ is defined by the formula
\begin{equation}
F_* M = D_{Y \leftarrow X} \otimes_{A_n} M.
\end{equation}

It is clearly an $A_m$-module. 

This is very convenient, because to compute direct images one has only to transpose the actions of $D_{X \to Y}$; and this is usually easy to do.

Let us work out the calculations for the projection $\pi: X \times Y \to Y$. 

From §1, we have that
\begin{equation}
D_{X \times Y \to Y} \cong A_{m+n} / \sum_{1}^{n} A_{m+n} \partial_{x_i}.
\end{equation}

Since the standard transposition of $A_{m+n}$ maps $\partial_{x_i}$ to $-\partial_{x_i}$, we conclude by Proposition 2.1 that

\begin{equation}
D_{Y \leftarrow X \times Y} \cong A_{m+n} / \sum_{1}^{n} \partial_{x_i} A_{m+n}.
\end{equation}

Note that the $y$'s commute with the $\partial_{x_i}$'s. 

In particular, $\sum_{1}^{n} \partial_{x_i} A_{m+n}$ is a left $A_m$-submodule of $A_{m+n}$. 

Thus $A_{m+n} / \sum_{1}^{n} \partial_{x_i} A_{m+n}$ is a left $A_m$-module for the quotient action.

The direct image under projections does {\color{red}not} take finitely generated modules to finitely generated modules. 

Indeed, $A_{m+n}$ is cyclic as a module over itself, but $A_{m+n} / \sum_{1}^{n} \partial_{x_i} A_{m+n}$ is not finitely generated over $A_m$. 

It is a free $A_m$-module; with a basis given by the monomials on the $x$'s.

Using this description of $D_{Y \leftarrow X \times Y}$ we can easily calculate the direct image of any module under a projection. 

If $M$ is a left $A_{m+n}$-module, then by Corollary 12.4.7,
\begin{equation}
\pi_* M \cong M / \sum_{1}^{n} \partial_{x_i} M.
\end{equation}

Another interesting example is provided by polynomial isomorphisms. 

Let $F: X \to Y$ be a polynomial map, and consider the polynomial isomorphism $G: X \times Y \to X \times Y$ defined by
\begin{equation}
G(X, Y) = (X, Y + F(X)).
\end{equation}

In Ch. 15, §3, we saw how to calculate inverse images under this map. 

In particular, there exists an automorphism $\sigma$ of $A_{m+n}$ such that
\begin{equation}
D_{X \times Y \to X \times Y} = G^*(A_{m+n}) \cong (A_{m+n})_\sigma.
\end{equation}

We wish to transpose this module in order to calculate direct images. 

The final result is very simple, but a technical lemma is required.

\textbf{Lemma.} 

Let $\tau$ be the standard transposition of $A_{m+n}$. 

Then $\tau \circ \sigma \circ \tau = \sigma$.

\textbf{Proof:} It is enough to check that the identity holds on the generators of $A_{m+n}$. 

First, $\tau$ is the identity on $K[X, Y]$ and $\sigma$ restricts to an automorphism of $K[X, Y]$. 

Thus $\tau \circ \sigma \circ \tau = \sigma$ in $K[X, Y]$. On the other hand,
\begin{equation}
\tau \sigma \tau(\partial_{x_i}) = \tau \sigma(-\partial_{x_i}) = -\tau(\partial_{x_i} + \sum_{1}^{m} \frac{\partial F_j}{\partial x_i} \partial_{y_j}).
\end{equation}

Since $\partial F_j / \partial x_i$ commutes with $\partial_{y_j}$ we have that
\begin{equation}
\tau \sigma \tau(\partial_{x_i}) = \sigma(\partial_{x_i}).
\end{equation}

Similarly, $\tau \sigma \tau(\partial_{y_j}) = \sigma(\partial_{y_j})$; and the proof is complete.


\textbf{Proposition.} 

Let $M$ be a left $A_{m+n}$-module, then
\begin{equation}
G_* M \cong M_{\sigma^{-1}}.
\end{equation}

\textbf{Proof:} Let us first find the transpose of $(A_{m+n})_\sigma$. 

Consider the map
\begin{equation}
\psi : ((A_{m+n})_\sigma)^t \to A_{m+n}
\end{equation}
defined by $\psi(u) = \tau(u)$. It is a $K$-vector space isomorphism. 

Furthermore, if $a, b \in A_{m+n}$ and $u \in ((A_{m+n})_\sigma)^t$ then
\begin{equation}
a \diamond u \diamond b = \tau(b) * u \tau(a) = \sigma(\tau(b)) u \tau(a).
\end{equation}

Therefore, $\psi(a \diamond u \diamond b) = a \tau(u) \tau \sigma \tau(b)$. 

By Lemma 3.1, $\tau \sigma \tau(b) = \sigma(b)$; and so
\begin{equation}
\psi(a \diamond u \diamond b) = a \tau(u) \sigma(b) = a \psi(u) * b.
\end{equation}

Hence $D_{X \times Y \leftarrow X \times Y}$ is $A_{m+n}$ as a bimodule over itself, but with the right action twisted by $\sigma$. 

Therefore,
\begin{equation}
G_* M \cong A_{m+n} \otimes_\sigma M.
\end{equation}

By Lemma 14.1.1 this is isomorphic to $M_{\sigma^{-1}}$.

\textbf{Corollary.} 

Let $M$ be a finitely generated left $A_{m+n}$-module. 

Then $G_* M$ and $M$ have the same dimension.

\textbf{Proof:} Follows from Proposition 3.2 and Corollary 9.2.4.


\end{frame}

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% Section 4
\section{Exercises}
\begin{frame}[allowframebreaks]{D. }


\textbf{Exercises 4.1.} Let $R$ be a $K$-algebra and $\tau_1$, $\tau_2$ two transpositions of $R$. 

Show that $\tau_1 (\tau_2)^{-1}$ is an automorphism of $R$.

\newpage

\textbf{Exercises 4.2.} Let $\pi: K^2 \to K$ be the projection on the second coordinate. 

Compute the direct image of the following $A_2$-modules under $\pi$.
\begin{enumerate}
    \item $A_2 / A_2 \partial_2$
    \item $K[x_1, x_2]$
    \item $A_2 / A_2 \partial_1$
    \item $A_2 / A_2 x_2$
    \item $A_2 / A_2 \partial_2^3$
\end{enumerate}


\newpage

\textbf{Exercises 4.3.} Which of the direct images of Exercise 4.2 are finitely generated over $A_1$?


\newpage

\textbf{Exercises 4.4.} Let $F: X \to Y$ be a polynomial map and $M$ a left $A_n$-module. Show that

\begin{equation}
F_* M \cong \left(M{\,}^t \otimes_{A_m} D_{X \to Y}\right)^t.
\end{equation}

\newpage

\textbf{Exercises 4.5.} Let $\iota: X \to X \times Y$ be the standard embedding and $\eta: X \times Y \to X$ be the projection on the first coordinate. 

Let $M$ be a left $A_{m+n}$-module. 

Show that $\iota^* M$ is the Fourier Transform of $\eta_* M$.



\newpage

\textbf{Exercises 4.6.} Let $F: X \to Y$ be a polynomial map. 

Define $G: X \times Y \to X \times Y$ by $G(X, Y) = (X, Y + F(X))$. 

Let $M$ be a left $A_{m+n}$-module. 

Show that
\begin{equation}
G_* G^* M = M = G^* G_* M.
\end{equation}


\end{frame}

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